Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 20

Table of contents
- a 20
  - Solution a
- b 20
  - Solution b

a

Solution

For the heat equation

Θ_{t} = κ Θ_{xx}

, separation of variables with homogeneous Neumann boundary condition at both ends gives spatial modes

cos (nπ x / L)

and decay factors

e^{- κ (nπ / L)^{2} t}

. The

n = 0

mode is constant and does not decay.

The initial profile already contains only two such modes:

T_{0} sin^{2} \frac{π x}{L} = \frac{T _{0}}{2} - \frac{T _{0}}{2} cos \frac{2 π x}{L} .

Evolving each mode therefore yields

Θ (x, t) = \frac{T _{0}}{2} [1 - e^{- 4 κ π^{2} t / L^{2}} cos \frac{2 π x}{L}] .

The endpoint derivatives vanish, and substitution checks both the equation and the initial profile. The spatial integral remains

L T_{0} /2

: no heat escapes through insulated ends, so the limiting uniform temperature is

T_{0} /2

.

Insulated and zero-temperature endpoints evolve the same initial heat profile differently

. Original mode calculations. Both panels start from the same squared sine. Insulated endpoints conserve the mean temperature and approach

T_{0} /2

; zero-temperature endpoints remove heat and the profile decays to zero. Times are expressed as

τ = κ π^{2} t / L^{2}

b

Solution

With homogeneous Dirichlet boundary condition, the eigenfunctions are

sin (nπ x / L)

and all modes decay. Write

Θ (x, t) = \sum_{n = 1}^{\infty} b_{n} sin \frac{nπ x}{L} e^{- κ (nπ / L)^{2} t}, b_{n} = \frac{2 T _{0}}{L} \int_{0}^{L} sin^{2} \frac{π x}{L} sin \frac{nπ x}{L} d x .

Reflection about the midpoint makes every even-

n

coefficient zero. For odd

n

, putting

u = π x / L

and using the product-to-sum identity gives

b_{n} = \frac{2 T _{0}}{π} \int_{0}^{π} sin^{2} u sin (n u) d u = \frac{2 T _{0}}{π} [\frac{1}{n} - \frac{1}{2} (\frac{1}{n + 2} + \frac{1}{n - 2})] = \frac{8 T _{0}}{πn ( 4 - n ^{2} )} .

In particular

b_{1} = 8 T_{0} / (3 π)

and

b_{3} = - 8 T_{0} / (15 π)

. Thus

Θ (x, t) = \frac{8 T _{0}}{π} m = 0 \sum \infty \frac{sin (( 2 m + 1 ) π x / L )}{( 2 m + 1 ) [ 4 - ( 2 m + 1 ) ^{2} ]} e^{- κ (2 m + 1)^{2} π^{2} t / L^{2}} .

This uses the Fourier sine series of a squared sine. The coefficients are

O (n^{- 3})

, so the initial series converges uniformly; for

t > 0

the exponential factors also justify termwise differentiation to check the heat equation. Every mode vanishes at both endpoints, and the initial series gives the specified profile. Unlike part (a), there is no constant mode: the temperature tends to zero, with leading long-time term

Θ (x, t) \sim \frac{8 T _{0}}{3 π} sin \frac{π x}{L} e^{- κ π^{2} t / L^{2}} .

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